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Integrate the following
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\[\int\limits_{}^{} \frac{ 4}{ \sqrt{3-4x^{2}} } dx\]
\[\int\frac{4}{\sqrt{3-4x^2}}dx\\ 4\int\frac{1}{\sqrt{3-4x^2}}dx\] Make the following substitution: \[x=\frac{\sqrt{3}}{2}\sin t\\ dx=\frac{\sqrt{3}}{2}\cos t\;dt\] The integral then becomes \[4\int\limits\frac{1}{\sqrt{3-4\left(\frac{\sqrt{3}}{2}\sin t\right)^2}}\left(\frac{\sqrt{3}}{2}\cos t\;dt\right)\]
Oh actually I figured it out after, what i was expected to do was to use arcsinx to solve. Thanks though!
Using this substitution will give you an answer in terms of arcsinx (after a few steps of simplification, of course), but I think there is some method of substitution directly involving arcsine. I'm just more familiar with this method. You're welcome.
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